Răspuns:
a; b; c ; dimensiunile paralelipipedului ; d²=a²+b²+c² ; unde d este diagonala ; a²+b²+c² = 100 ; a/3=b/4=c/5=k ; a=3k ; b=4k . c=5k ; 9k²+16k²+25k²=100 ; 50k²=100 ; k²=2 ' k=√2 ; a=3√2 ; b=4 √2 ; c=5√2 ; At=2ab+2ac+2bc= 2(3√2×4√2+3√2×5√2+4√2×5√2)=2(24+30+40)=94×2=198 ; V=a.b.c=3√2.4√2.5√2= 60.2√2=120√2 cm³;
Explicație pas cu pas:
Fie a, b, c dimensiunile paralelipipedului și d, lungimea diagonalei.
[tex]\it d^2=a^2+b^2+c^2=100\ \ \ \ \ (1)\\ \\ \{a,b,c\}\ d.\ p.\ \{3,\ 4,\ 5\}\Rightarrow \dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5} \Rightarrow \dfrac{a^2}{9}=\dfrac{b^2}{16}=\dfrac{c^2}{25}=\\ \\ \\ =\dfrac{a^2+b^2+c^2}{9+16+25}\stackrel{(1)}{=}\ \dfrac{100}{50}=2 \Rightarrow \begin{cases}\it \dfrac{a^2}{9}=2 \Rightarrow a^2=9\cdot2\\ \\\it \dfrac{b^2}{16}=2 \Rightarrow b^2=16\cdot2\\ \\ \it \dfrac{c^2}{25}=2 \Rightarrow c^2=25\cdot2\end{cases}\ \ \ \ (2)[/tex]
[tex]\it \mathcal{A}_t =2(ab+bc+ca);\ \ \ \ \mathcal{V}=abc\\ \\ \\(2) \Rightarrow \begin{cases} \it a^2b^2=9\cdot2\cdot16\cdot2=9\cdot64\Rightarrow ab=3\cdot8=24\\ \\ \it b^2c^2=16\cdot2\cdot25\cdot2=16\cdot100\Rightarrow bc=4\cdot10=40\\ \\ \it c^2a^2=25\cdot2\cdot9\cdot2=9\cdot100\Rightarrow ca=3\cdot10=30\end{cases} \ \ \ \ (3)[/tex]
[tex]\it \mathcal{A}_t=2(ab+bc+ca) \stackrel{(3)}{\Longrightarrow} \mathcal{A}_t=2(24+40+30) =2\cdot94 =188\ cm^2\\ \\ (2) \Rightarrow a^2b^2c^2=9\cdot2\cdot16\cdot2\cdot25\cdot2=9\cdot16\cdot100\cdot2 \Rightarrow abc=3\cdot4\cdot10\sqrt2 =120\sqrt2\\ \\ \mathcal{V} =abc=120\sqrt2\ cm^3[/tex]